Facility Location with Service Installation

Costs

Chaitanya Swamy

Joint work with David Shmoys and Retsef Levi

Cornell University

SODA Talk, 01/2004

Facility Location

F : set of facilities.D : set of clients.

Facility i has facility cost fi.

cij : distance between points i and j.

client

facility

SODA Talk, 01/2004

Facility Locationwith Services

F : set of facilities.D : set of clients.S : set of services={tan, green,

pink}.

Facility i has facility cost fi.

cij : distance between points i and j.

Client j requires service s(j).fi,s : cost of installing service s at

fac. i.

facility

client

services

SODA Talk, 01/2004

2) Install a set of services S(i) on each open facility i.

We want to:

Cost = ∑iA fi + ∑iA, sS(i) fi,s + ∑jD ci(j)j

= facility opening cost + service installation cost + client assignment cost

facility

client

services

3) Assign each client j to an open facility i(j) on which service s(j) is installed.

1) Choose a set A of facilities to open.

open facility

SODA Talk, 01/2004

Related Work

Only one service uncapacitated facility location• LP rounding: Shmoys, Tardos &

Aardal; Chudak & Shmoys; Sviridenko.

• Primal-dual algorithms: Jain & Vazirani; Jain, Mahdian, Markakis, Saberi & Vazirani.

Best approx. - Mahdian, Ye & Zhang: 1.52

Baev & Rajaraman consider related caching problem: No facility costs, capacity on the number of services that may be installed at a facility. Gave a 20.5-approx. algorithm, improved to 10 by (S).

Ravi & Sinha consider similar model – multicommodity facility location. Give a log(|S|)-approx. algorithm.

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Our Results

• Give a 6-approx. primal-dual algorithm. Assume facilities can be ordered s.t. if i ≤ i’, then fi,s ≤ fi’,s for all services s. Special cases: fi,s = hs – depends only on s

fi,s = gi – depends only on i

General problem is set-cover hard.

• When fi,s = hs primal-dual gives a 5-approx. Improve ratio to 2.391 using LP rounding.

• Consider k-median variant: give a 11.6-approx. when fi,s = hs using primal-dual+Lagrangian relaxation.

SODA Talk, 01/2004

LP formulation

Minimize ∑i fiyi+∑i,s fi,syi,s+∑j,i cijxij

(Primal)

subject to∑i xij ≥ 1 j

xij ≤ yi,s(j) i, j

xij ≤ yi i, j

xij, yi ≥ 0 i, j

yi : indicates if facility i is open.

yi,s : indicates if service s is installed on facility i.

xij : indicates if client j is assigned to facility i.

s(j) : service required by j.

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∑i,jD(s) zij ≤ fi,s i, s

The Dual

D(s) = clients requiring service s

∑i,j wij ≤ fi i

cij

zij

wij

vj, wij, zij ≥ 0 i, j

Maximize ∑j vj

subject tovj ≤ cij + wij + zij i, j

vj

j

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Algorithm Outline

Based on the primal-dual method.

Any feasible dual solution is a lower bound on OPT - Weak Duality.1) Construct primal solution and

dual solution simultaneously.

2) Bound cost of primal by c●(dual value) get a c-approx.

Algorithm strongly motivated by the Jain-Vazirani algorithm.

Notion of time, t. Initially, t=0, all dual variables vj, zij, wij are 0. No facility is open, no service is installed.

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fd,s = 2 for all services s.

fc,green = 2.

t=0

unfrozen client

not openfacility

a

c

d

b

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Raise all vj at rate 1.

t=0

unfrozen client

not openfacility

a

c

d

b

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t=1

unfrozen client

not openfacility

fd,s = 2 for all services s.

fc,green = 2.

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t=2

unfrozen client

not openfacility

fd,s = 2 for all services s.

fc,green = 2.

If vj ≥ cij say that j is tight with i.

j becomes tight with i and s(j) is not installed on i: start increasing zij.

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t=3

unfrozen client

not openfacility

fd,s = 2 for all services s.

fc,green = 2.

For some i, service s, ∑i,jD(s) zij = fi,s: tentatively install s on i, start raising wij for all j in D(s) tight with i.

SODA Talk, 01/2004

unfrozen client

not openfacility

t=4

unfrozen client

not openfacilitytentativelyopen facility

frozenclient

fd,s = 2 for all services s.

fc,green = 2.

For some i, ∑i,j wij = fi : tentatively open i, if j is tight with i and s(j) is tentatively installed at i, freeze j.

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t=5

unfrozen client

not openfacilitytentativelyopen facility

frozenclient

fd,s = 2 for all services s.

fc,green = 2.

When all demands are frozen, process stops.

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The Primal-Dual process

Keep raising all vj at rate 1 until:

1) j reaches facility i: if s(j) is not tentatively installed at i, increase zij; else, if i is not tentatively open, raise wij; otherwise freeze j.

2) For some i and s, ∑i,jD(s) zij = fi,s: tentatively install s on i. If i is not tentatively open raise wij for all j in D(s) tight with i, else freeze j.

3) For some i, ∑i,j wij = fi : tentatively open i. If j is tight with i and s(j) is tentatively installed at i, freeze j.

Now only raise vj of unfrozen demands. Continue until all demands are frozen.

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Opening facilities

a

c

d

b

X

X

Say i, i’ are dependent if there is client j s.t. wij, wi’j > 0.

Suppose O is a ≤ b ≤ c ≤ d. Consider tentatively open facilities in this order, pick a maximal independent subset.

Open all these facilities – call this set F’.

Let O: ordering on facilities.

i tentatively open, not opened there is i’ s.t. i’ ≤ i and i, i’ are dependent.

Denote i’ = nbr(i).

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Installing services

X

X F’ = facilities opened.

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X

X

Installing services

Consider service s. Let Fs = tentatively open facilities where service s is tentatively installed.

Say i, i’ are s-dependent if there is a client jD(s) s.t. zij, zi’j > 0. Carefully pick a maximal independent set F’s Fs.

F’ = facilities opened.

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X

X

Installing services

For each i in F’s, if iF’ install service s on i, otherwise install s on nbr(i).

F’ = facilities opened.

Consider service s. Let Fs = tentatively open facilities where service s is tentatively installed.

Say i, i’ are s-dependent if there is a client jD(s) s.t. zij, zi’j > 0. Carefully pick a maximal independent set F’s Fs.

SODA Talk, 01/2004

Installing services

For each i in F’s, if iF’ install service s on i, otherwise install s on nbr(i).

F’ = facilities opened.

Consider service s. Let Fs = tentatively open facilities where service s is tentatively installed.

Say i, i’ are s-dependent if there is a client jD(s) s.t. zij, zi’j > 0. Carefully pick a maximal independent set F’s Fs.

X

X

SODA Talk, 01/2004

Installing services

For each i in F’s, if iF’ install service s on i, otherwise install s on nbr(i).

F’ = facilities opened.

Consider service s. Let Fs = tentatively open facilities where service s is tentatively installed.

Say i, i’ are s-dependent if there is a client jD(s) s.t. zij, zi’j > 0. Carefully pick a maximal independent set F’s Fs.

X

X

SODA Talk, 01/2004

Installing services

F’ = facilities opened.

For each i in F’s, if iF’ install service s on i, otherwise install s on nbr(i).

Consider service s. Let Fs = tentatively open facilities where service s is tentatively installed.

Say i, i’ are s-dependent if there is a client jD(s) s.t. zij, zi’j > 0. Carefully pick a maximal independent set F’s Fs.

X

X

SODA Talk, 01/2004

X

X

Installing services

For each i in F’s, if iF’ install service s on i, otherwise install s on nbr(i).

F’ = facilities opened.

Consider service s. Let Fs = tentatively open facilities where service s is tentatively installed.

Say i, i’ are s-dependent if there is a client jD(s) s.t. zij, zi’j > 0. Carefully pick a maximal independent set F’s Fs.

SODA Talk, 01/2004

X

X

Installing services

For each i in F’s, if iF’ install service s on i, otherwise install s on nbr(i).

F’ = facilities opened.

Consider service s. Let Fs = tentatively open facilities where service s is tentatively installed.

Say i, i’ are s-dependent if there is a client jD(s) s.t. zij, zi’j > 0. Carefully pick a maximal independent set F’s Fs.

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Finally,

Client j is assigned to the nearest open facility on which service s(j) is installed.

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Analysis sketch

Let D’ = {j: i in F’ s.t. wij > 0}.

Facility opening cost ≤ ∑jD’ vj.

Service installation cost ≤ ∑j vj.

Can show that the assignment cost of j is at most 3vj if jD’ and at most 5vj otherwise.

SODA Talk, 01/2004

Analysis sketch

Let D’ = {j: i in F’ s.t. wij > 0}.

Facility opening cost ≤ ∑jD’ vj.

Service installation cost ≤ ∑j vj.

Can show that the assignment cost of j is at most 3vj if jD’ and at most 5vj otherwise.

Theorem: Total cost ≤ 6●∑j vj ≤ 6●OPT.

SODA Talk, 01/2004

Thank You.